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• Specific issue on data pump API in oracle

  - by Median Hilal

  I have a client/server architecture. Using an Oracle dbms on the database server side. I need to perform a user-triggered (from client side) backup of the database, where the best way to perform that is using a stored procedure on the server side which the client may call, as the client has no oracle tools to perform the backup. I've searched thorough inside available solutions and have found that using a stored procedure is the best way. Well, then I found that using oracle data pump API is the best way to use inside a PL/SQl stored procedure. My specific questions about the API are... I would like to ask about two issues ... ---- The first ----- the detach function to detach the handler, is it necessary to be used at the end of the procedure? and what if I don't use it? I read the Oracle documentation but I didn't get their point, they say it doesn't terminate the job but indicates that the user is not interested in it, an when I use detach at the end of my procedure the exported .dmp file disappears. ---- The second ----- to perform a user (client side) triggered back up as the modification are only to the data, I used TABLE parameter for the export operation. But the version parameter... what should it be? I also read the documentation but couldn't determine what I need (LATEST or COMPATIBLE) ? Thanks

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• Amtrak's Mobile-SOA Oracle SOA Solution at OpenWorld

  - by Bruce Tierney

  During yesterday's Mobile SOA Session, Innowave presented their ticketing solution implemented for Amtrak which uses Oracle SOA Suite for service-enablement with support for Microsoft Windows Mobile handheld devices.  Innowave's Hilal Khan described this chart and highlighted the value of a service-based approach since the data went to handhelds as well as to APEX reports with a single service implementation:

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• “if” statement vs OO Design - 2

  - by hilal

  I encountered similar problem “if” statement vs OO Design - 1 but it is slightly different. Here is the problem that open the popup (different objects/popups) onValueChange of listbox Popup1 p1; // different objects Popup2 p2; // different objects Popup3 p3; ... listbox.add("p1"); listbox.add("p2"); listbox.add("p3"); ... listbox.addChangeHandler() { if(getSelectedItem().equals("p1")){ p1 = new Popup1(); p1.show(); } else if() {...} .... } I don't want to write "if" that if p1 then p1 = new Popup1(); p1.center(); How I can handle this situation? Any design-pattern? Here is my solution but it is so costly map() { map.put("p1", new Popup1()); map.put("p2", new Popup2()); map.put("p3", new Popup3()); } onValueChange() { map.get(selectedItem).show(); } One drawback is initialization all the popups. but it is require only when valueChange

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• Intelligent web features, algorithms (people you may follow, similar to you ...)

  - by hilal

  I have 3 main questions about the algorithms in intelligent web (web 2.0) Here the book I'm reading http://www.amazon.com/Algorithms-Intelligent-Web-Haralambos-Marmanis/dp/1933988665 and I want to learn the algorithms in deeper 1. People You may follow (Twitter) How can one determine the nearest result to my requests ? Data mining? which algorithms? 2. How you’re connected feature (Linkedin) Simply algorithm works like that. It draws the path between two nodes let say between Me and the other person is C. Me - A, B - A connections - C . It is not any brute force algorithms or any other like graph algorithms :) 3. Similar to you (Twitter, Facebook) This algorithms is similar to 1. Does it simply work the max(count) friend in common (facebook) or the max(count) follower in Twitter? or any other algorithms they implement? I think the second part is true because running the loop dict{count, person} for person in contacts: dict.add(count(common(person))) return dict(max) is a silly act in every refreshing page. 4. Did you mean (Google) I know that they may implement it with phonetic algorithm http://en.wikipedia.org/wiki/Phonetic_algorithm simply soundex http://en.wikipedia.org/wiki/Soundex and here is the Google VP of Engineering and CIO Douglas Merrill speak http://www.youtube.com/watch?v=syKY8CrHkck#t=22m03s What about first 3 questions? Any ideas are welcome ! Thanks

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• why Cannot invoke super constructor from enum constructor ?

  - by hilal

  public enum A { A(1); private A(int i){ } private A(){ super(); // compile - error // Cannot invoke super constructor from enum constructor A() } } and here is the hierarchy of enum A extends from abstract java.lang.Enum extends java.lang.Object Class c = Class.forName("/*path*/.A"); System.out.println(c.getSuperclass().getName()); System.out.println(Modifier.toString(c.getSuperclass().getModifiers()).contains("abstract")); System.out.println(c.getSuperclass().getSuperclass().getName());

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• minimum L sum in a mxn matrix - 2

  - by hilal

  Here is my first question about maximum L sum and here is different and hard version of it. Problem : Given a mxn *positive* integer matrix find the minimum L sum from 0th row to the m'th row . L(4 item) likes chess horse move Example : M = 3x3 0 1 2 1 3 2 4 2 1 Possible L moves are : (0 1 2 2), (0 1 3 2) (0 1 4 2) We should go from 0th row to the 3th row with minimum sum I solved this with dynamic-programming and here is my algorithm : 1. Take a mxn another Minimum L Moves Sum array and copy the first row of main matrix. I call it (MLMS) 2. start from first cell and look the up L moves and calculate it 3. insert it in MLMS if it is less than exists value 4. Do step 2. until m'th row 5. Choose the minimum sum in the m'th row Let me explain on my example step by step: M[ 0 ][ 0 ] sum(L1 = (0, 1, 2, 2)) = 5 ; sum(L2 = (0,1,3,2)) = 6; so MLMS[ 0 ][ 1 ] = 6 sum(L3 = (0, 1, 3, 2)) = 6 ; sum(L4 = (0,1,4,2)) = 7; so MLMS[ 2 ][ 1 ] = 6 M[ 0 ][ 1 ] sum(L5 = (1, 0, 1, 4)) = 6; sum(L6 = (1,3,2,4)) = 10; so MLMS[ 2 ][ 2 ] = 6 ... the last MSLS is : 0 1 2 4 3 6 6 6 6 Which means 6 is the minimum L sum that can be reach from 0 to the m. I think it is O(8*(m-1)*n) = O(m*n). Is there any optimal solution or dynamic-programming algorithms fit this problem? Thanks, sorry for long question

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• computes the number of possible orderings of n objects under the relations < and =

  - by hilal

  Here is the problem : Give a algorithm that takes a positive integer n as input, and computes the number of possible orderings of n objects under the relations < and =. For example, if n = 3 the 13 possible orderings are as follows: a = b = c, a = b < c, a < b = c, a < b < c, a < c < b, a = c < b, b < a = c, b < a < c, b < c < a, b = c < a, c < a = b, c < a < b, c < b < a. Your algorithm should run in time polynomial in n. I'm null to this problem. Can you find any solution to this dynamic-programming problem?

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• Given a number N, find the number of ways to write it as a sum of two or more consecutive integers

  - by hilal

  Here is the problem (Given a number N, find the number of ways to write it as a sum of two or more consecutive integers) and example 15 = 7+8, 1+2+3+4+5, 4+5+6 I solved with math like that : a + (a + 1) + (a + 2) + (a + 3) + ... + (a + k) = N (k + 1)*a + (1 + 2 + 3 + ... + k) = N (k + 1)a + k(k+1)/2 = N (k + 1)*(2*a + k)/2 = N Then check that if N divisible by (k+1) and (2*a+k) then I can find answer in O(N) time Here is my question how can you solve this by dynamic-programming ? and what is the complexity (O) ? P.S : excuse me, if it is a duplicate question. I searched but I can find

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• Interview question : What is the fastest way to generate prime number recursively ?

  - by hilal

  Generation of prime number is simple but what is the fastest way to find it and generate( prime numbers) it recursively ? Here is my solution. However, it is not the best way. I think it is O(N*sqrt(N)). Please correct me, if I am wrong. public static boolean isPrime(int n) { if (n < 2) { return false; } else if (n % 2 == 0 & n != 2) { return false; } else { return isPrime(n, (int) Math.sqrt(n)); } } private static boolean isPrime(int n, int i) { if (i < 2) { return true; } else if (n % i == 0) { return false; } else { return isPrime(n, --i); } } public static void generatePrimes(int n){ if(n < 2) { return ; } else if(isPrime(n)) { System.out.println(n); } generatePrimes(--n); } public static void main(String[] args) { generatePrimes(200); }

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• How&rsquo;s your Momma an&rsquo; them?

  - by Bill Jones Jr.

  When a Southern “boy” like me sees somebody that used to be, or should be, a close friend or relative that they haven’t seen in a long time, that’s a typical greeting.  Come to think of it, we were often related to close friends. So “back in the day”, we not only knew people but everybody close to them.  When I started driving, my Dad told me to always drive carefully in Polk county.  He said if I ran into anybody there, it was likely they would be related or close family friends. Not so much any more… the cities have gotten bigger and more people come south and stay.  One of the curses of air conditioning I guess. Anyway, it’s been a while.  So “How’s your Momma and them”?  Have you been waiting for me to blog again?  Too bad, I’m back anyway <smile>. Here in Charlotte we just had another great code camp.  The Enterprise Developers Guild is going strong, thanks to the help of a lot of dedicated people.  Mark Wilson, Brian Gough, Syl Walker, Ghayth Hilal, Alberto Botero, Dan Thyer, Jean Doiron, Matt Duffield all come to mind.  Plus all the regulars who volunteer for every special event we have. Brian Gough put on a successful SharePoint Saturday.  Rafael Salas and our friends at the local Pass SQL group had a great SQL Saturday.  Brian Hitney and Glen Gordon keep on doing their usual great job for developers in the southeast as our local Microsoft reps. Since my last post, I have the honor of being designated the INetA Membership Mentor for Georgia in addition to mentoring the groups in the Carolinas for the past several years.  Georgia could be a really good thing since my wife likes shopping in Atlanta, not to mention how much we both like Georgia in general.  As I recall, my Momma had people in Georgia.  Wonder how their “Mommas an’ them” are doing?   Bill J

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